Wish to evaluate a generic function f\\((t\\)). Suppose we know a starting value, for example, f(0) = \\(c\\). We also know that we can express f\\((\\cdot\\)) through the ordinary differential equation
\n\\begin{eqnarray*}
\n\\frac{\\partial}{\\partial t} \\textrm{f}(t) = \\textrm{g}(\\textrm{f}(t)),
\n\\end{eqnarray*}
\nwhere \\(\\textrm{g}(\\cdot)\\) is a known function. Occasionally, we can solve the ordinary differential equation analytically. More often, a numerical solution is available. The easiest to explain and to implement is Euler’s method. To see this, consider a small \\(h\\) and use a straight line approximation for the derivative
\n\\begin{eqnarray*}
\n\\frac{\\partial}{\\partial t} \\textrm{f}(t) \\approx \\frac{\\textrm{f}(t+h)-\\textrm{f}(t)}{h}.
\n\\end{eqnarray*}
\nWith this, we have an approximation to the ordinary differential equation
\n\\begin{eqnarray*}
\n\\textrm{f}(t+h) \\approx \\textrm{f}(t) + h \\textrm{g}(\\textrm{f}(t))
\n\\end{eqnarray*}
\nthat we can evaluate recursively. <\/p>\n
Example – Interest Theory<\/strong> For this example, we can evaluate f(\\(\\cdot\\)) analytically, in that
\nSuppose that \\(\\textrm{g}(t) = 0.05 t\\), a line, and f(0) = 1. Then, if \\(h=0.001\\), we have
\n\\begin{eqnarray*}
\n\\textrm{f}(t+h) \\approx \\textrm{f}(t) + (0.001) (0.05) \\textrm{f}(t) .
\n\\end{eqnarray*}<\/p>\n\n
\n t<\/td>\n Approximate f(t<\/i>)<\/td>\n Exact f(t<\/i>)<\/td>\n Error <\/td>\n<\/tr>\n \n 0<\/td>\n 1.0000000000<\/td>\n 1.0000000000<\/td>\n 0.0000000000<\/td>\n<\/tr>\n \n 0.001<\/td>\n 1.0000500000<\/td>\n 1.0000500013<\/td>\n 0.0000000013<\/td>\n<\/tr>\n \n 0.002<\/td>\n 1.0001000025<\/td>\n 1.0001000050<\/td>\n 0.0000000025<\/td>\n<\/tr>\n \n 0.998<\/td>\n 1.0511646632<\/td>\n 1.0511659745<\/td>\n 0.0000013113<\/td>\n<\/tr>\n \n 0.999<\/td>\n 1.0512172215<\/td>\n 1.0512185341<\/td>\n 0.0000013127<\/td>\n<\/tr>\n \n 1<\/td>\n 1.0512697823<\/td>\n 1.0512710964<\/td>\n 0.0000013140<\/td>\n<\/tr>\n<\/table>\n
\n\\begin{eqnarray*}
\n\\frac{\\partial}{\\partial t} \\textrm{f}(t) = 0.05 \\textrm{f}(t) \\ \\ \\ \\ \\Rightarrow \\ \\ \\ \\ \\frac{\\partial}{\\partial t} \\ln \\textrm{f}(t) = 0.05 ,
\n\\end{eqnarray*}
\nand so that, with f(0) = 1, we have \\( \\textrm{f}(t) = e^{0.05 t}\\). <\/p>\n